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A solution initially has a [Ca2+] = 0.100 M and [Ba2+] = 0.100 M Sulfate ion (SO42-) is being added to selectively precipitate out the Ba2+ ions and leave behind the Ca2+ ions. (The barium ions will start to precipitate out first because barium sulfate is less soluble than calcium sulfate.) BaSO4 Ksp = 1.07 x 10-10 CaSO4 Ksp = 7.10 x 10-5 Assume that somehow the sulfate ion is being added in a manner that does not change the volume of the solution significantly.

a) At what concentration of sulfate ion would the Ca2+ begin to precipitate out as calcium sulfate?

b) At that concentration of sulfate ion what would be the concentration of Ba2+ ion?

Sagot :

Mickymike92idn

The concentration of sulfate ion would be required for Ca⁺ to begin to precipitate out as calcium sulfate is 7.10 * 10⁻⁴ M.

b) The concentration of the Ba²⁺ ion would be 9.929 * 10⁻² M

What concentration of sulfate ion would be required for Ca⁺ to begin to precipitate out as calcium sulfate?

The precipitation of calcium sulfate occurs when the ionic product, Kip is greater than or equal to the solubility product, Ksp of calcium sulfate.

  • Kip ≥ Ksp  

a) The equation of the dissociation of CaSO₄ is given below:

CaSO₄ ⇄ Ca²⁺ + SO₄²⁻

Ksp = [Ca²⁺] * [SO₄²⁻] = 7.10 * 10⁻⁵

Kip = [Ca²⁺] * [SO₄²⁻]

[SO₄²⁻] = Kip/[Ca²⁺]

[Ca²⁺] = 0.100 M

For precipitation to occur;

Kip = Ksp = 7.10 * 10⁻⁵

[SO₄²⁻] = 7.10 * 10⁻⁵/0.100

[SO₄²⁻] = 7.10 * 10⁻⁴ M

Therefore, the concentration of sulfate ion would be required for Ca⁺ to begin to precipitate out as calcium sulfate is 7.10 * 10⁻⁴ M.

b) The concentration of the Ba²⁺ ion would be;

0.100 M - 7.10 * 10⁻⁴ M = 9.929 * 10⁻² M

Learn more about solubility product at: https://brainly.com/question/20372961

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